Discussion
Consider \(\mathbb{R}^{n}\) with the usual topology for \(n = 1, 2, 3\). Each of the following options gives topological spaces \(X\) and \(Y\) with respective induced topologies. In which option is \(X\) homeomorphic to \(Y\)?
Answer: A
\(X\) is homeomorphic to \(Y\) in the first option: the cylinder \(x^{2} + y^{2} = 1\) in \(\mathbb{R}^{3}\) and the punctured plane \(z = 0\), \(x^{2} + y^{2} \neq 0\).
The cylinder is \(S^{1} \times \mathbb{R}\). In polar form the punctured plane is \(S^{1} \times (0, \infty)\), and \((0, \infty)\) is homeomorphic to \(\mathbb{R}\) by \(r \mapsto \ln r\). Explicitly, \((\cos\theta, \sin\theta, z) \mapsto (e^{z}\cos\theta, e^{z}\sin\theta, 0)\) is a homeomorphism.
The other pairs differ in a topological property. In the second, \(X\) is the topologist's sine curve, which is connected but not path connected, while \([0, 1]\) is path connected. In the third, \(X\) is the graph of a continuous function on \((0, 1]\), so it is homeomorphic to \((0, 1]\), which is not compact, while \([0, 1]\) is. In the fourth, the cone \(x^{2} + y^{2} = z^{2}\) without its vertex splits into the parts \(z \gt 0\) and \(z \lt 0\), so it is disconnected, while the cylinder is connected.
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